Question prev next
Solve the system:
\[y = - x - 8\]
\[y = \frac{x}{3} - 4\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = - \frac{4 x}{3} - 4\]
Subtract -4 from both sides.
\[4 = - \frac{4 x}{3}\]
Multiply/divide to solve for \(x\) .
\[x = -3\]
Plug in -3 for \(x\) in either original equation.
\[y = - (-3) - 8\]
\[y = -5\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = 3 - 3 x\]
\[y = 2 x + 8\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = - 5 x - 5\]
Subtract -5 from both sides.
\[5 = - 5 x\]
Multiply/divide to solve for \(x\) .
\[x = -1\]
Plug in -1 for \(x\) in either original equation.
\[y = 3 - 3 (-1)\]
\[y = 6\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = 7 x + 8\]
\[y = x - 4\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = 6 x + 12\]
Subtract 12 from both sides.
\[-12 = 6 x\]
Multiply/divide to solve for \(x\) .
\[x = -2\]
Plug in -2 for \(x\) in either original equation.
\[y = 7 (-2) + 8\]
\[y = -6\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = 4 x + 5\]
\[y = 2 x + 1\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = 2 x + 4\]
Subtract 4 from both sides.
\[-4 = 2 x\]
Multiply/divide to solve for \(x\) .
\[x = -2\]
Plug in -2 for \(x\) in either original equation.
\[y = 4 (-2) + 5\]
\[y = -3\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = 2 - 2 x\]
\[y = 7 - 3 x\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = x - 5\]
Subtract -5 from both sides.
\[5 = x\]
Multiply/divide to solve for \(x\) .
\[x = 5\]
Plug in 5 for \(x\) in either original equation.
\[y = 2 - 2 (5)\]
\[y = -8\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = - x - 8\]
\[y = x + 4\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = - 2 x - 12\]
Subtract -12 from both sides.
\[12 = - 2 x\]
Multiply/divide to solve for \(x\) .
\[x = -6\]
Plug in -6 for \(x\) in either original equation.
\[y = - (-6) - 8\]
\[y = -2\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = x + 6\]
\[y = - \frac{8 x}{5} - 7\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = \frac{13 x}{5} + 13\]
Subtract 13 from both sides.
\[-13 = \frac{13 x}{5}\]
Multiply/divide to solve for \(x\) .
\[x = -5\]
Plug in -5 for \(x\) in either original equation.
\[y = (-5) + 6\]
\[y = 1\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = 7 - x\]
\[y = - 7 x - 5\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = 6 x + 12\]
Subtract 12 from both sides.
\[-12 = 6 x\]
Multiply/divide to solve for \(x\) .
\[x = -2\]
Plug in -2 for \(x\) in either original equation.
\[y = 7 - (-2)\]
\[y = 9\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = x - 7\]
\[y = 2 x - 4\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = - x - 3\]
Subtract -3 from both sides.
\[3 = - x\]
Multiply/divide to solve for \(x\) .
\[x = -3\]
Plug in -3 for \(x\) in either original equation.
\[y = (-3) - 7\]
\[y = -10\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = 8 - \frac{5 x}{2}\]
\[y = 6 - 2 x\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = 2 - \frac{x}{2}\]
Subtract 2 from both sides.
\[-2 = - \frac{x}{2}\]
Multiply/divide to solve for \(x\) .
\[x = 4\]
Plug in 4 for \(x\) in either original equation.
\[y = 8 - \frac{5 (4)}{2}\]
\[y = -2\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = - \frac{x}{3} - 2\]
\[y = x + 2\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = - \frac{4 x}{3} - 4\]
Subtract -4 from both sides.
\[4 = - \frac{4 x}{3}\]
Multiply/divide to solve for \(x\) .
\[x = -3\]
Plug in -3 for \(x\) in either original equation.
\[y = - \frac{(-3)}{3} - 2\]
\[y = -1\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = - x - 4\]
\[y = x + 6\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = - 2 x - 10\]
Subtract -10 from both sides.
\[10 = - 2 x\]
Multiply/divide to solve for \(x\) .
\[x = -5\]
Plug in -5 for \(x\) in either original equation.
\[y = - (-5) - 4\]
\[y = 1\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = 8 - x\]
\[y = x + 4\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = 4 - 2 x\]
Subtract 4 from both sides.
\[-4 = - 2 x\]
Multiply/divide to solve for \(x\) .
\[x = 2\]
Plug in 2 for \(x\) in either original equation.
\[y = 8 - (2)\]
\[y = 6\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = \frac{7 x}{5} - 4\]
\[y = 8 - x\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = \frac{12 x}{5} - 12\]
Subtract -12 from both sides.
\[12 = \frac{12 x}{5}\]
Multiply/divide to solve for \(x\) .
\[x = 5\]
Plug in 5 for \(x\) in either original equation.
\[y = \frac{7 (5)}{5} - 4\]
\[y = 3\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = 2 - 2 x\]
\[y = - \frac{3 x}{2} - 1\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = 3 - \frac{x}{2}\]
Subtract 3 from both sides.
\[-3 = - \frac{x}{2}\]
Multiply/divide to solve for \(x\) .
\[x = 6\]
Plug in 6 for \(x\) in either original equation.
\[y = 2 - 2 (6)\]
\[y = -10\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = 5 - x\]
\[y = 3 - \frac{3 x}{2}\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = \frac{x}{2} + 2\]
Subtract 2 from both sides.
\[-2 = \frac{x}{2}\]
Multiply/divide to solve for \(x\) .
\[x = -4\]
Plug in -4 for \(x\) in either original equation.
\[y = 5 - (-4)\]
\[y = 9\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = x - 1\]
\[y = - \frac{2 x}{5} - 8\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = \frac{7 x}{5} + 7\]
Subtract 7 from both sides.
\[-7 = \frac{7 x}{5}\]
Multiply/divide to solve for \(x\) .
\[x = -5\]
Plug in -5 for \(x\) in either original equation.
\[y = (-5) - 1\]
\[y = -6\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = 3 x + 8\]
\[y = 2 x + 7\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = x + 1\]
Subtract 1 from both sides.
\[-1 = x\]
Multiply/divide to solve for \(x\) .
\[x = -1\]
Plug in -1 for \(x\) in either original equation.
\[y = 3 (-1) + 8\]
\[y = 5\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = 2 x + 4\]
\[y = 4 x + 2\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = 2 - 2 x\]
Subtract 2 from both sides.
\[-2 = - 2 x\]
Multiply/divide to solve for \(x\) .
\[x = 1\]
Plug in 1 for \(x\) in either original equation.
\[y = 2 (1) + 4\]
\[y = 6\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[y = 4 - x\]
\[y = 2 - 2 x\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
Eliminate \(y\) by subtrating the second equation from the first.
\[0 = x + 2\]
Subtract 2 from both sides.
\[-2 = x\]
Multiply/divide to solve for \(x\) .
\[x = -2\]
Plug in -2 for \(x\) in either original equation.
\[y = 4 - (-2)\]
\[y = 6\]
You can also solve this by graphing!
Question prev next
Solve the system:
\[- 6 x + 5 y = 30\]
\[x - 2 y = 2\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[-6x_{\text{int}1}+(5)(0) = 30 \]
\[-6x_{\text{int}1} = 30 \]
\[x_{\text{int}1} = \frac{30}{-6} = -5 \]
\[1x_{\text{int}2}+(-2)(0) = 2 \]
\[1x_{\text{int}2} = 2 \]
\[x_{\text{int}2} = \frac{2}{1} = 2 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(-6)(0)+(5)(y_{\text{int1}}) = 30 \]
\[5y_{\text{int}1} = 30 \]
\[y_{\text{int}1} = \frac{30}{5} = 6 \]
\[(1)(0)+(-2)y_{\text{int2}} = 2 \]
\[-2y_{\text{int}2} = 2 \]
\[y_{\text{int}2} = \frac{2}{-2} = -1 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((-10,\,-6)\) .
Question prev next
Solve the system:
\[- x + y = 9\]
\[- x - 2 y = 6\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[-1x_{\text{int}1}+(1)(0) = 9 \]
\[-1x_{\text{int}1} = 9 \]
\[x_{\text{int}1} = \frac{9}{-1} = -9 \]
\[-1x_{\text{int}2}+(-2)(0) = 6 \]
\[-1x_{\text{int}2} = 6 \]
\[x_{\text{int}2} = \frac{6}{-1} = -6 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(-1)(0)+(1)(y_{\text{int1}}) = 9 \]
\[1y_{\text{int}1} = 9 \]
\[y_{\text{int}1} = \frac{9}{1} = 9 \]
\[(-1)(0)+(-2)y_{\text{int2}} = 6 \]
\[-2y_{\text{int}2} = 6 \]
\[y_{\text{int}2} = \frac{6}{-2} = -3 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((-8,\,1)\) .
Question prev next
Solve the system:
\[3 x - 7 y = 21\]
\[- x + y = 1\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[3x_{\text{int}1}+(-7)(0) = 21 \]
\[3x_{\text{int}1} = 21 \]
\[x_{\text{int}1} = \frac{21}{3} = 7 \]
\[-1x_{\text{int}2}+(1)(0) = 1 \]
\[-1x_{\text{int}2} = 1 \]
\[x_{\text{int}2} = \frac{1}{-1} = -1 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(3)(0)+(-7)(y_{\text{int1}}) = 21 \]
\[-7y_{\text{int}1} = 21 \]
\[y_{\text{int}1} = \frac{21}{-7} = -3 \]
\[(-1)(0)+(1)y_{\text{int2}} = 1 \]
\[1y_{\text{int}2} = 1 \]
\[y_{\text{int}2} = \frac{1}{1} = 1 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((-7,\,-6)\) .
Question prev next
Solve the system:
\[4 x - 3 y = 24\]
\[x - y = 5\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[4x_{\text{int}1}+(-3)(0) = 24 \]
\[4x_{\text{int}1} = 24 \]
\[x_{\text{int}1} = \frac{24}{4} = 6 \]
\[1x_{\text{int}2}+(-1)(0) = 5 \]
\[1x_{\text{int}2} = 5 \]
\[x_{\text{int}2} = \frac{5}{1} = 5 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(4)(0)+(-3)(y_{\text{int1}}) = 24 \]
\[-3y_{\text{int}1} = 24 \]
\[y_{\text{int}1} = \frac{24}{-3} = -8 \]
\[(1)(0)+(-1)y_{\text{int2}} = 5 \]
\[-1y_{\text{int}2} = 5 \]
\[y_{\text{int}2} = \frac{5}{-1} = -5 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((9,\,4)\) .
Question prev next
Solve the system:
\[3 x + 2 y = 6\]
\[- 3 x + 2 y = 18\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[3x_{\text{int}1}+(2)(0) = 6 \]
\[3x_{\text{int}1} = 6 \]
\[x_{\text{int}1} = \frac{6}{3} = 2 \]
\[-3x_{\text{int}2}+(2)(0) = 18 \]
\[-3x_{\text{int}2} = 18 \]
\[x_{\text{int}2} = \frac{18}{-3} = -6 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(3)(0)+(2)(y_{\text{int1}}) = 6 \]
\[2y_{\text{int}1} = 6 \]
\[y_{\text{int}1} = \frac{6}{2} = 3 \]
\[(-3)(0)+(2)y_{\text{int2}} = 18 \]
\[2y_{\text{int}2} = 18 \]
\[y_{\text{int}2} = \frac{18}{2} = 9 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((-2,\,6)\) .
Question prev next
Solve the system:
\[- 3 x + 2 y = 12\]
\[- 3 x - y = 3\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[-3x_{\text{int}1}+(2)(0) = 12 \]
\[-3x_{\text{int}1} = 12 \]
\[x_{\text{int}1} = \frac{12}{-3} = -4 \]
\[-3x_{\text{int}2}+(-1)(0) = 3 \]
\[-3x_{\text{int}2} = 3 \]
\[x_{\text{int}2} = \frac{3}{-3} = -1 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(-3)(0)+(2)(y_{\text{int1}}) = 12 \]
\[2y_{\text{int}1} = 12 \]
\[y_{\text{int}1} = \frac{12}{2} = 6 \]
\[(-3)(0)+(-1)y_{\text{int2}} = 3 \]
\[-1y_{\text{int}2} = 3 \]
\[y_{\text{int}2} = \frac{3}{-1} = -3 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((-2,\,3)\) .
Question prev next
Solve the system:
\[x - y = 3\]
\[x - 2 y = 4\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[1x_{\text{int}1}+(-1)(0) = 3 \]
\[1x_{\text{int}1} = 3 \]
\[x_{\text{int}1} = \frac{3}{1} = 3 \]
\[1x_{\text{int}2}+(-2)(0) = 4 \]
\[1x_{\text{int}2} = 4 \]
\[x_{\text{int}2} = \frac{4}{1} = 4 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(1)(0)+(-1)(y_{\text{int1}}) = 3 \]
\[-1y_{\text{int}1} = 3 \]
\[y_{\text{int}1} = \frac{3}{-1} = -3 \]
\[(1)(0)+(-2)y_{\text{int2}} = 4 \]
\[-2y_{\text{int}2} = 4 \]
\[y_{\text{int}2} = \frac{4}{-2} = -2 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((2,\,-1)\) .
Question prev next
Solve the system:
\[- x + 2 y = 8\]
\[- x + y = 3\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[-1x_{\text{int}1}+(2)(0) = 8 \]
\[-1x_{\text{int}1} = 8 \]
\[x_{\text{int}1} = \frac{8}{-1} = -8 \]
\[-1x_{\text{int}2}+(1)(0) = 3 \]
\[-1x_{\text{int}2} = 3 \]
\[x_{\text{int}2} = \frac{3}{-1} = -3 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(-1)(0)+(2)(y_{\text{int1}}) = 8 \]
\[2y_{\text{int}1} = 8 \]
\[y_{\text{int}1} = \frac{8}{2} = 4 \]
\[(-1)(0)+(1)y_{\text{int2}} = 3 \]
\[1y_{\text{int}2} = 3 \]
\[y_{\text{int}2} = \frac{3}{1} = 3 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((2,\,5)\) .
Question prev next
Solve the system:
\[- 3 x + y = 9\]
\[- 3 x - 2 y = 18\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[-3x_{\text{int}1}+(1)(0) = 9 \]
\[-3x_{\text{int}1} = 9 \]
\[x_{\text{int}1} = \frac{9}{-3} = -3 \]
\[-3x_{\text{int}2}+(-2)(0) = 18 \]
\[-3x_{\text{int}2} = 18 \]
\[x_{\text{int}2} = \frac{18}{-3} = -6 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(-3)(0)+(1)(y_{\text{int1}}) = 9 \]
\[1y_{\text{int}1} = 9 \]
\[y_{\text{int}1} = \frac{9}{1} = 9 \]
\[(-3)(0)+(-2)y_{\text{int2}} = 18 \]
\[-2y_{\text{int}2} = 18 \]
\[y_{\text{int}2} = \frac{18}{-2} = -9 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((-4,\,-3)\) .
Question prev next
Solve the system:
\[- 3 x - y = 9\]
\[x - 2 y = 4\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[-3x_{\text{int}1}+(-1)(0) = 9 \]
\[-3x_{\text{int}1} = 9 \]
\[x_{\text{int}1} = \frac{9}{-3} = -3 \]
\[1x_{\text{int}2}+(-2)(0) = 4 \]
\[1x_{\text{int}2} = 4 \]
\[x_{\text{int}2} = \frac{4}{1} = 4 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(-3)(0)+(-1)(y_{\text{int1}}) = 9 \]
\[-1y_{\text{int}1} = 9 \]
\[y_{\text{int}1} = \frac{9}{-1} = -9 \]
\[(1)(0)+(-2)y_{\text{int2}} = 4 \]
\[-2y_{\text{int}2} = 4 \]
\[y_{\text{int}2} = \frac{4}{-2} = -2 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((-2,\,-3)\) .
Question prev next
Solve the system:
\[- 2 x + 3 y = 18\]
\[- x + y = 8\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[-2x_{\text{int}1}+(3)(0) = 18 \]
\[-2x_{\text{int}1} = 18 \]
\[x_{\text{int}1} = \frac{18}{-2} = -9 \]
\[-1x_{\text{int}2}+(1)(0) = 8 \]
\[-1x_{\text{int}2} = 8 \]
\[x_{\text{int}2} = \frac{8}{-1} = -8 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(-2)(0)+(3)(y_{\text{int1}}) = 18 \]
\[3y_{\text{int}1} = 18 \]
\[y_{\text{int}1} = \frac{18}{3} = 6 \]
\[(-1)(0)+(1)y_{\text{int2}} = 8 \]
\[1y_{\text{int}2} = 8 \]
\[y_{\text{int}2} = \frac{8}{1} = 8 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((-6,\,2)\) .
Question prev next
Solve the system:
\[- x + 2 y = 2\]
\[- x + 4 y = 8\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[-1x_{\text{int}1}+(2)(0) = 2 \]
\[-1x_{\text{int}1} = 2 \]
\[x_{\text{int}1} = \frac{2}{-1} = -2 \]
\[-1x_{\text{int}2}+(4)(0) = 8 \]
\[-1x_{\text{int}2} = 8 \]
\[x_{\text{int}2} = \frac{8}{-1} = -8 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(-1)(0)+(2)(y_{\text{int1}}) = 2 \]
\[2y_{\text{int}1} = 2 \]
\[y_{\text{int}1} = \frac{2}{2} = 1 \]
\[(-1)(0)+(4)y_{\text{int2}} = 8 \]
\[4y_{\text{int}2} = 8 \]
\[y_{\text{int}2} = \frac{8}{4} = 2 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((4,\,3)\) .
Question prev next
Solve the system:
\[x - y = 1\]
\[- x - 2 y = 8\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[1x_{\text{int}1}+(-1)(0) = 1 \]
\[1x_{\text{int}1} = 1 \]
\[x_{\text{int}1} = \frac{1}{1} = 1 \]
\[-1x_{\text{int}2}+(-2)(0) = 8 \]
\[-1x_{\text{int}2} = 8 \]
\[x_{\text{int}2} = \frac{8}{-1} = -8 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(1)(0)+(-1)(y_{\text{int1}}) = 1 \]
\[-1y_{\text{int}1} = 1 \]
\[y_{\text{int}1} = \frac{1}{-1} = -1 \]
\[(-1)(0)+(-2)y_{\text{int2}} = 8 \]
\[-2y_{\text{int}2} = 8 \]
\[y_{\text{int}2} = \frac{8}{-2} = -4 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((-2,\,-3)\) .
Question prev next
Solve the system:
\[x - y = 7\]
\[2 x - y = 8\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[1x_{\text{int}1}+(-1)(0) = 7 \]
\[1x_{\text{int}1} = 7 \]
\[x_{\text{int}1} = \frac{7}{1} = 7 \]
\[2x_{\text{int}2}+(-1)(0) = 8 \]
\[2x_{\text{int}2} = 8 \]
\[x_{\text{int}2} = \frac{8}{2} = 4 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(1)(0)+(-1)(y_{\text{int1}}) = 7 \]
\[-1y_{\text{int}1} = 7 \]
\[y_{\text{int}1} = \frac{7}{-1} = -7 \]
\[(2)(0)+(-1)y_{\text{int2}} = 8 \]
\[-1y_{\text{int}2} = 8 \]
\[y_{\text{int}2} = \frac{8}{-1} = -8 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((1,\,-6)\) .
Question prev next
Solve the system:
\[- 3 x - y = 3\]
\[3 x + 2 y = 6\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[-3x_{\text{int}1}+(-1)(0) = 3 \]
\[-3x_{\text{int}1} = 3 \]
\[x_{\text{int}1} = \frac{3}{-3} = -1 \]
\[3x_{\text{int}2}+(2)(0) = 6 \]
\[3x_{\text{int}2} = 6 \]
\[x_{\text{int}2} = \frac{6}{3} = 2 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(-3)(0)+(-1)(y_{\text{int1}}) = 3 \]
\[-1y_{\text{int}1} = 3 \]
\[y_{\text{int}1} = \frac{3}{-1} = -3 \]
\[(3)(0)+(2)y_{\text{int2}} = 6 \]
\[2y_{\text{int}2} = 6 \]
\[y_{\text{int}2} = \frac{6}{2} = 3 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((-4,\,9)\) .
Question prev next
Solve the system:
\[- 2 x + y = 6\]
\[- x + y = 2\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[-2x_{\text{int}1}+(1)(0) = 6 \]
\[-2x_{\text{int}1} = 6 \]
\[x_{\text{int}1} = \frac{6}{-2} = -3 \]
\[-1x_{\text{int}2}+(1)(0) = 2 \]
\[-1x_{\text{int}2} = 2 \]
\[x_{\text{int}2} = \frac{2}{-1} = -2 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(-2)(0)+(1)(y_{\text{int1}}) = 6 \]
\[1y_{\text{int}1} = 6 \]
\[y_{\text{int}1} = \frac{6}{1} = 6 \]
\[(-1)(0)+(1)y_{\text{int2}} = 2 \]
\[1y_{\text{int}2} = 2 \]
\[y_{\text{int}2} = \frac{2}{1} = 2 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((-4,\,-2)\) .
Question prev next
Solve the system:
\[2 x + y = 2\]
\[x - y = 7\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[2x_{\text{int}1}+(1)(0) = 2 \]
\[2x_{\text{int}1} = 2 \]
\[x_{\text{int}1} = \frac{2}{2} = 1 \]
\[1x_{\text{int}2}+(-1)(0) = 7 \]
\[1x_{\text{int}2} = 7 \]
\[x_{\text{int}2} = \frac{7}{1} = 7 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(2)(0)+(1)(y_{\text{int1}}) = 2 \]
\[1y_{\text{int}1} = 2 \]
\[y_{\text{int}1} = \frac{2}{1} = 2 \]
\[(1)(0)+(-1)y_{\text{int2}} = 7 \]
\[-1y_{\text{int}2} = 7 \]
\[y_{\text{int}2} = \frac{7}{-1} = -7 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((3,\,-4)\) .
Question prev next
Solve the system:
\[x - y = 1\]
\[- 4 x + y = 8\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[1x_{\text{int}1}+(-1)(0) = 1 \]
\[1x_{\text{int}1} = 1 \]
\[x_{\text{int}1} = \frac{1}{1} = 1 \]
\[-4x_{\text{int}2}+(1)(0) = 8 \]
\[-4x_{\text{int}2} = 8 \]
\[x_{\text{int}2} = \frac{8}{-4} = -2 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(1)(0)+(-1)(y_{\text{int1}}) = 1 \]
\[-1y_{\text{int}1} = 1 \]
\[y_{\text{int}1} = \frac{1}{-1} = -1 \]
\[(-4)(0)+(1)y_{\text{int2}} = 8 \]
\[1y_{\text{int}2} = 8 \]
\[y_{\text{int}2} = \frac{8}{1} = 8 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((-3,\,-4)\) .
Question prev next
Solve the system:
\[4 x - 5 y = 20\]
\[x - 2 y = 2\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[4x_{\text{int}1}+(-5)(0) = 20 \]
\[4x_{\text{int}1} = 20 \]
\[x_{\text{int}1} = \frac{20}{4} = 5 \]
\[1x_{\text{int}2}+(-2)(0) = 2 \]
\[1x_{\text{int}2} = 2 \]
\[x_{\text{int}2} = \frac{2}{1} = 2 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(4)(0)+(-5)(y_{\text{int1}}) = 20 \]
\[-5y_{\text{int}1} = 20 \]
\[y_{\text{int}1} = \frac{20}{-5} = -4 \]
\[(1)(0)+(-2)y_{\text{int2}} = 2 \]
\[-2y_{\text{int}2} = 2 \]
\[y_{\text{int}2} = \frac{2}{-2} = -1 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((10,\,4)\) .
Question prev next
Solve the system:
\[- 2 x + 5 y = 10\]
\[6 x - 5 y = 30\]
Your answers:
\(x=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 \(y=\) -------- -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solution Show solution
When graphing linear equations given in standard form, it is typically best to determine both the \(x\) -intercept and the \(y\) -intercept.
To find the \(x\) -intercepts, set \(y\) to 0.
\[-2x_{\text{int}1}+(5)(0) = 10 \]
\[-2x_{\text{int}1} = 10 \]
\[x_{\text{int}1} = \frac{10}{-2} = -5 \]
\[6x_{\text{int}2}+(-5)(0) = 30 \]
\[6x_{\text{int}2} = 30 \]
\[x_{\text{int}2} = \frac{30}{6} = 5 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(-2)(0)+(5)(y_{\text{int1}}) = 10 \]
\[5y_{\text{int}1} = 10 \]
\[y_{\text{int}1} = \frac{10}{5} = 2 \]
\[(6)(0)+(-5)y_{\text{int2}} = 30 \]
\[-5y_{\text{int}2} = 30 \]
\[y_{\text{int}2} = \frac{30}{-5} = -6 \]
Plot those intercepts, determine slope pattern, draw lines, and find intersection.
The intersection occurs at \((10,\,6)\) .
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 5 kilograms and a volume of 4 liters. Each yiv has a mass of 15 kilograms and a volume of 18 liters. The thief can carry a maximum mass of 75 kilograms and a maximum volume of 72 liters. The profit from each xot is $1.8 and the profit from each yiv is $7.1.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
5
15
75
volume (L)
4
18
72
profit ($)
1.80
7.10
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(5x+15y \le 75\)
Volume constraint: \(4x+18y \le 72\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[5x+15y = 75\]
\[4x+18y = 72\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[1x+3y = 15\]
\[2x+9y = 36\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[1x_{\text{int}1}+(3)(0) = 15 \]
\[1x_{\text{int}1} = 15 \]
\[x_{\text{int}1} = \frac{15}{1} = 15 \]
\[2x_{\text{int}2}+(9)(0) = 36 \]
\[2x_{\text{int}2} = 36 \]
\[x_{\text{int}2} = \frac{36}{2} = 18 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(1)(0)+(3)(y_{\text{int1}}) = 15 \]
\[3y_{\text{int}1} = 15 \]
\[y_{\text{int}1} = \frac{15}{3} = 5 \]
\[(2)(0)+(9)y_{\text{int2}} = 36 \]
\[9y_{\text{int}2} = 36 \]
\[y_{\text{int}2} = \frac{36}{9} = 4 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[3x + 9y = 45\]
\[2x + 9y = 36\]
Subtract equations.
\[1x + (0)y = 9\]
Simplify.
\[1x = 9\]
Divide by 1.
\[x = 9\]
Plug \(x = 9\) into the first equation. Solve for \(y\) .
\[(1)(9)+3y = 15\]
\[9+3y = 15\]
\[3y = 6\]
\[y = 2\]
So, the intersection occurs at \((9,\,2)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,4)\) or \((9,2)\) or \((15,0)\) .
Find those profits, using \(P(x,y)=1.8x+7.1y\) .
\[P(0,4)=(1.8)(0)+(7.1)(4) = 28.4\]
\[P(9,2)=(1.8)(9)+(7.1)(2) = 30.4\]
\[P(15,0)=(1.8)(15)+(7.1)(0) = 27\]
So, the maximum profit is $30.40.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 10 kilograms and a volume of 18 liters. Each yiv has a mass of 20 kilograms and a volume of 12 liters. The thief can carry a maximum mass of 200 kilograms and a maximum volume of 216 liters. The profit from each xot is $6.5 and the profit from each yiv is $4.8.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
10
20
200
volume (L)
18
12
216
profit ($)
6.50
4.80
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(10x+20y \le 200\)
Volume constraint: \(18x+12y \le 216\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[10x+20y = 200\]
\[18x+12y = 216\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[1x+2y = 20\]
\[3x+2y = 36\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[1x_{\text{int}1}+(2)(0) = 20 \]
\[1x_{\text{int}1} = 20 \]
\[x_{\text{int}1} = \frac{20}{1} = 20 \]
\[3x_{\text{int}2}+(2)(0) = 36 \]
\[3x_{\text{int}2} = 36 \]
\[x_{\text{int}2} = \frac{36}{3} = 12 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(1)(0)+(2)(y_{\text{int1}}) = 20 \]
\[2y_{\text{int}1} = 20 \]
\[y_{\text{int}1} = \frac{20}{2} = 10 \]
\[(3)(0)+(2)y_{\text{int2}} = 36 \]
\[2y_{\text{int}2} = 36 \]
\[y_{\text{int}2} = \frac{36}{2} = 18 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[1x + 2y = 20\]
\[3x + 2y = 36\]
Subtract equations.
\[2x + (0)y = 16\]
Simplify.
\[2x = 16\]
Divide by 2.
\[x = 8\]
Plug \(x = 8\) into the first equation. Solve for \(y\) .
\[(1)(8)+2y = 20\]
\[8+2y = 20\]
\[2y = 12\]
\[y = 6\]
So, the intersection occurs at \((8,\,6)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,10)\) or \((8,6)\) or \((12,0)\) .
Find those profits, using \(P(x,y)=6.5x+4.8y\) .
\[P(0,10)=(6.5)(0)+(4.8)(10) = 48\]
\[P(8,6)=(6.5)(8)+(4.8)(6) = 80.8\]
\[P(12,0)=(6.5)(12)+(4.8)(0) = 78\]
So, the maximum profit is $80.80.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 6 kilograms and a volume of 8 liters. Each yiv has a mass of 3 kilograms and a volume of 2 liters. The thief can carry a maximum mass of 18 kilograms and a maximum volume of 16 liters. The profit from each xot is $9.53 and the profit from each yiv is $3.7.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
6
3
18
volume (L)
8
2
16
profit ($)
9.53
3.70
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(6x+3y \le 18\)
Volume constraint: \(8x+2y \le 16\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[6x+3y = 18\]
\[8x+2y = 16\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[2x+1y = 6\]
\[4x+1y = 8\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[2x_{\text{int}1}+(1)(0) = 6 \]
\[2x_{\text{int}1} = 6 \]
\[x_{\text{int}1} = \frac{6}{2} = 3 \]
\[4x_{\text{int}2}+(1)(0) = 8 \]
\[4x_{\text{int}2} = 8 \]
\[x_{\text{int}2} = \frac{8}{4} = 2 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(2)(0)+(1)(y_{\text{int1}}) = 6 \]
\[1y_{\text{int}1} = 6 \]
\[y_{\text{int}1} = \frac{6}{1} = 6 \]
\[(4)(0)+(1)y_{\text{int2}} = 8 \]
\[1y_{\text{int}2} = 8 \]
\[y_{\text{int}2} = \frac{8}{1} = 8 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[2x + 1y = 6\]
\[4x + 1y = 8\]
Subtract equations.
\[2x + (0)y = 2\]
Simplify.
\[2x = 2\]
Divide by 2.
\[x = 1\]
Plug \(x = 1\) into the first equation. Solve for \(y\) .
\[(2)(1)+1y = 6\]
\[2+1y = 6\]
\[1y = 4\]
\[y = 4\]
So, the intersection occurs at \((1,\,4)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,6)\) or \((1,4)\) or \((2,0)\) .
Find those profits, using \(P(x,y)=9.53x+3.7y\) .
\[P(0,6)=(9.53)(0)+(3.7)(6) = 22.2\]
\[P(1,4)=(9.53)(1)+(3.7)(4) = 24.33\]
\[P(2,0)=(9.53)(2)+(3.7)(0) = 19.06\]
So, the maximum profit is $24.33.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 7 kilograms and a volume of 4 liters. Each yiv has a mass of 14 kilograms and a volume of 20 liters. The thief can carry a maximum mass of 98 kilograms and a maximum volume of 80 liters. The profit from each xot is $8.15 and the profit from each yiv is $3.12.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
7
14
98
volume (L)
4
20
80
profit ($)
8.15
3.12
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(7x+14y \le 98\)
Volume constraint: \(4x+20y \le 80\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[7x+14y = 98\]
\[4x+20y = 80\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[1x+2y = 14\]
\[1x+5y = 20\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[1x_{\text{int}1}+(2)(0) = 14 \]
\[1x_{\text{int}1} = 14 \]
\[x_{\text{int}1} = \frac{14}{1} = 14 \]
\[1x_{\text{int}2}+(5)(0) = 20 \]
\[1x_{\text{int}2} = 20 \]
\[x_{\text{int}2} = \frac{20}{1} = 20 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(1)(0)+(2)(y_{\text{int1}}) = 14 \]
\[2y_{\text{int}1} = 14 \]
\[y_{\text{int}1} = \frac{14}{2} = 7 \]
\[(1)(0)+(5)y_{\text{int2}} = 20 \]
\[5y_{\text{int}2} = 20 \]
\[y_{\text{int}2} = \frac{20}{5} = 4 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[5x + 10y = 70\]
\[2x + 10y = 40\]
Subtract equations.
\[3x + (0)y = 30\]
Simplify.
\[3x = 30\]
Divide by 3.
\[x = 10\]
Plug \(x = 10\) into the first equation. Solve for \(y\) .
\[(1)(10)+2y = 14\]
\[10+2y = 14\]
\[2y = 4\]
\[y = 2\]
So, the intersection occurs at \((10,\,2)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,4)\) or \((10,2)\) or \((14,0)\) .
Find those profits, using \(P(x,y)=8.15x+3.12y\) .
\[P(0,4)=(8.15)(0)+(3.12)(4) = 12.48\]
\[P(10,2)=(8.15)(10)+(3.12)(2) = 87.74\]
\[P(14,0)=(8.15)(14)+(3.12)(0) = 114.1\]
So, the maximum profit is $114.10.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 8 kilograms and a volume of 18 liters. Each yiv has a mass of 16 kilograms and a volume of 6 liters. The thief can carry a maximum mass of 128 kilograms and a maximum volume of 108 liters. The profit from each xot is $6.9 and the profit from each yiv is $3.29.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
8
16
128
volume (L)
18
6
108
profit ($)
6.90
3.29
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(8x+16y \le 128\)
Volume constraint: \(18x+6y \le 108\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[8x+16y = 128\]
\[18x+6y = 108\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[1x+2y = 16\]
\[3x+1y = 18\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[1x_{\text{int}1}+(2)(0) = 16 \]
\[1x_{\text{int}1} = 16 \]
\[x_{\text{int}1} = \frac{16}{1} = 16 \]
\[3x_{\text{int}2}+(1)(0) = 18 \]
\[3x_{\text{int}2} = 18 \]
\[x_{\text{int}2} = \frac{18}{3} = 6 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(1)(0)+(2)(y_{\text{int1}}) = 16 \]
\[2y_{\text{int}1} = 16 \]
\[y_{\text{int}1} = \frac{16}{2} = 8 \]
\[(3)(0)+(1)y_{\text{int2}} = 18 \]
\[1y_{\text{int}2} = 18 \]
\[y_{\text{int}2} = \frac{18}{1} = 18 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[1x + 2y = 16\]
\[6x + 2y = 36\]
Subtract equations.
\[5x + (0)y = 20\]
Simplify.
\[5x = 20\]
Divide by 5.
\[x = 4\]
Plug \(x = 4\) into the first equation. Solve for \(y\) .
\[(1)(4)+2y = 16\]
\[4+2y = 16\]
\[2y = 12\]
\[y = 6\]
So, the intersection occurs at \((4,\,6)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,8)\) or \((4,6)\) or \((6,0)\) .
Find those profits, using \(P(x,y)=6.9x+3.29y\) .
\[P(0,8)=(6.9)(0)+(3.29)(8) = 26.32\]
\[P(4,6)=(6.9)(4)+(3.29)(6) = 47.34\]
\[P(6,0)=(6.9)(6)+(3.29)(0) = 41.4\]
So, the maximum profit is $47.34.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 10 kilograms and a volume of 7 liters. Each yiv has a mass of 8 kilograms and a volume of 14 liters. The thief can carry a maximum mass of 80 kilograms and a maximum volume of 98 liters. The profit from each xot is $5.72 and the profit from each yiv is $5.52.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
10
8
80
volume (L)
7
14
98
profit ($)
5.72
5.52
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(10x+8y \le 80\)
Volume constraint: \(7x+14y \le 98\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[10x+8y = 80\]
\[7x+14y = 98\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[5x+4y = 40\]
\[1x+2y = 14\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[5x_{\text{int}1}+(4)(0) = 40 \]
\[5x_{\text{int}1} = 40 \]
\[x_{\text{int}1} = \frac{40}{5} = 8 \]
\[1x_{\text{int}2}+(2)(0) = 14 \]
\[1x_{\text{int}2} = 14 \]
\[x_{\text{int}2} = \frac{14}{1} = 14 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(5)(0)+(4)(y_{\text{int1}}) = 40 \]
\[4y_{\text{int}1} = 40 \]
\[y_{\text{int}1} = \frac{40}{4} = 10 \]
\[(1)(0)+(2)y_{\text{int2}} = 14 \]
\[2y_{\text{int}2} = 14 \]
\[y_{\text{int}2} = \frac{14}{2} = 7 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[5x + 4y = 40\]
\[2x + 4y = 28\]
Subtract equations.
\[3x + (0)y = 12\]
Simplify.
\[3x = 12\]
Divide by 3.
\[x = 4\]
Plug \(x = 4\) into the first equation. Solve for \(y\) .
\[(5)(4)+4y = 40\]
\[20+4y = 40\]
\[4y = 20\]
\[y = 5\]
So, the intersection occurs at \((4,\,5)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,7)\) or \((4,5)\) or \((8,0)\) .
Find those profits, using \(P(x,y)=5.72x+5.52y\) .
\[P(0,7)=(5.72)(0)+(5.52)(7) = 38.64\]
\[P(4,5)=(5.72)(4)+(5.52)(5) = 50.48\]
\[P(8,0)=(5.72)(8)+(5.52)(0) = 45.76\]
So, the maximum profit is $50.48.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 8 kilograms and a volume of 5 liters. Each yiv has a mass of 12 kilograms and a volume of 15 liters. The thief can carry a maximum mass of 96 kilograms and a maximum volume of 75 liters. The profit from each xot is $3.82 and the profit from each yiv is $7.91.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
8
12
96
volume (L)
5
15
75
profit ($)
3.82
7.91
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(8x+12y \le 96\)
Volume constraint: \(5x+15y \le 75\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[8x+12y = 96\]
\[5x+15y = 75\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[2x+3y = 24\]
\[1x+3y = 15\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[2x_{\text{int}1}+(3)(0) = 24 \]
\[2x_{\text{int}1} = 24 \]
\[x_{\text{int}1} = \frac{24}{2} = 12 \]
\[1x_{\text{int}2}+(3)(0) = 15 \]
\[1x_{\text{int}2} = 15 \]
\[x_{\text{int}2} = \frac{15}{1} = 15 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(2)(0)+(3)(y_{\text{int1}}) = 24 \]
\[3y_{\text{int}1} = 24 \]
\[y_{\text{int}1} = \frac{24}{3} = 8 \]
\[(1)(0)+(3)y_{\text{int2}} = 15 \]
\[3y_{\text{int}2} = 15 \]
\[y_{\text{int}2} = \frac{15}{3} = 5 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[2x + 3y = 24\]
\[1x + 3y = 15\]
Subtract equations.
\[1x + (0)y = 9\]
Simplify.
\[1x = 9\]
Divide by 1.
\[x = 9\]
Plug \(x = 9\) into the first equation. Solve for \(y\) .
\[(2)(9)+3y = 24\]
\[18+3y = 24\]
\[3y = 6\]
\[y = 2\]
So, the intersection occurs at \((9,\,2)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,5)\) or \((9,2)\) or \((12,0)\) .
Find those profits, using \(P(x,y)=3.82x+7.91y\) .
\[P(0,5)=(3.82)(0)+(7.91)(5) = 39.55\]
\[P(9,2)=(3.82)(9)+(7.91)(2) = 50.2\]
\[P(12,0)=(3.82)(12)+(7.91)(0) = 45.84\]
So, the maximum profit is $50.20.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 12 kilograms and a volume of 3 liters. Each yiv has a mass of 6 kilograms and a volume of 15 liters. The thief can carry a maximum mass of 72 kilograms and a maximum volume of 45 liters. The profit from each xot is $6.31 and the profit from each yiv is $4.25.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
12
6
72
volume (L)
3
15
45
profit ($)
6.31
4.25
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(12x+6y \le 72\)
Volume constraint: \(3x+15y \le 45\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[12x+6y = 72\]
\[3x+15y = 45\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[2x+1y = 12\]
\[1x+5y = 15\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[2x_{\text{int}1}+(1)(0) = 12 \]
\[2x_{\text{int}1} = 12 \]
\[x_{\text{int}1} = \frac{12}{2} = 6 \]
\[1x_{\text{int}2}+(5)(0) = 15 \]
\[1x_{\text{int}2} = 15 \]
\[x_{\text{int}2} = \frac{15}{1} = 15 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(2)(0)+(1)(y_{\text{int1}}) = 12 \]
\[1y_{\text{int}1} = 12 \]
\[y_{\text{int}1} = \frac{12}{1} = 12 \]
\[(1)(0)+(5)y_{\text{int2}} = 15 \]
\[5y_{\text{int}2} = 15 \]
\[y_{\text{int}2} = \frac{15}{5} = 3 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[10x + 5y = 60\]
\[1x + 5y = 15\]
Subtract equations.
\[9x + (0)y = 45\]
Simplify.
\[9x = 45\]
Divide by 9.
\[x = 5\]
Plug \(x = 5\) into the first equation. Solve for \(y\) .
\[(2)(5)+1y = 12\]
\[10+1y = 12\]
\[1y = 2\]
\[y = 2\]
So, the intersection occurs at \((5,\,2)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,3)\) or \((5,2)\) or \((6,0)\) .
Find those profits, using \(P(x,y)=6.31x+4.25y\) .
\[P(0,3)=(6.31)(0)+(4.25)(3) = 12.75\]
\[P(5,2)=(6.31)(5)+(4.25)(2) = 40.05\]
\[P(6,0)=(6.31)(6)+(4.25)(0) = 37.86\]
So, the maximum profit is $40.05.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 9 kilograms and a volume of 12 liters. Each yiv has a mass of 18 kilograms and a volume of 6 liters. The thief can carry a maximum mass of 162 kilograms and a maximum volume of 72 liters. The profit from each xot is $5.78 and the profit from each yiv is $4.67.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
9
18
162
volume (L)
12
6
72
profit ($)
5.78
4.67
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(9x+18y \le 162\)
Volume constraint: \(12x+6y \le 72\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[9x+18y = 162\]
\[12x+6y = 72\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[1x+2y = 18\]
\[2x+1y = 12\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[1x_{\text{int}1}+(2)(0) = 18 \]
\[1x_{\text{int}1} = 18 \]
\[x_{\text{int}1} = \frac{18}{1} = 18 \]
\[2x_{\text{int}2}+(1)(0) = 12 \]
\[2x_{\text{int}2} = 12 \]
\[x_{\text{int}2} = \frac{12}{2} = 6 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(1)(0)+(2)(y_{\text{int1}}) = 18 \]
\[2y_{\text{int}1} = 18 \]
\[y_{\text{int}1} = \frac{18}{2} = 9 \]
\[(2)(0)+(1)y_{\text{int2}} = 12 \]
\[1y_{\text{int}2} = 12 \]
\[y_{\text{int}2} = \frac{12}{1} = 12 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[1x + 2y = 18\]
\[4x + 2y = 24\]
Subtract equations.
\[3x + (0)y = 6\]
Simplify.
\[3x = 6\]
Divide by 3.
\[x = 2\]
Plug \(x = 2\) into the first equation. Solve for \(y\) .
\[(1)(2)+2y = 18\]
\[2+2y = 18\]
\[2y = 16\]
\[y = 8\]
So, the intersection occurs at \((2,\,8)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,9)\) or \((2,8)\) or \((6,0)\) .
Find those profits, using \(P(x,y)=5.78x+4.67y\) .
\[P(0,9)=(5.78)(0)+(4.67)(9) = 42.03\]
\[P(2,8)=(5.78)(2)+(4.67)(8) = 48.92\]
\[P(6,0)=(5.78)(6)+(4.67)(0) = 34.68\]
So, the maximum profit is $48.92.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 12 kilograms and a volume of 20 liters. Each yiv has a mass of 9 kilograms and a volume of 5 liters. The thief can carry a maximum mass of 108 kilograms and a maximum volume of 100 liters. The profit from each xot is $8.6 and the profit from each yiv is $7.11.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
12
9
108
volume (L)
20
5
100
profit ($)
8.60
7.11
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(12x+9y \le 108\)
Volume constraint: \(20x+5y \le 100\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[12x+9y = 108\]
\[20x+5y = 100\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[4x+3y = 36\]
\[4x+1y = 20\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[4x_{\text{int}1}+(3)(0) = 36 \]
\[4x_{\text{int}1} = 36 \]
\[x_{\text{int}1} = \frac{36}{4} = 9 \]
\[4x_{\text{int}2}+(1)(0) = 20 \]
\[4x_{\text{int}2} = 20 \]
\[x_{\text{int}2} = \frac{20}{4} = 5 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(4)(0)+(3)(y_{\text{int1}}) = 36 \]
\[3y_{\text{int}1} = 36 \]
\[y_{\text{int}1} = \frac{36}{3} = 12 \]
\[(4)(0)+(1)y_{\text{int2}} = 20 \]
\[1y_{\text{int}2} = 20 \]
\[y_{\text{int}2} = \frac{20}{1} = 20 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[4x + 3y = 36\]
\[12x + 3y = 60\]
Subtract equations.
\[8x + (0)y = 24\]
Simplify.
\[8x = 24\]
Divide by 8.
\[x = 3\]
Plug \(x = 3\) into the first equation. Solve for \(y\) .
\[(4)(3)+3y = 36\]
\[12+3y = 36\]
\[3y = 24\]
\[y = 8\]
So, the intersection occurs at \((3,\,8)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,12)\) or \((3,8)\) or \((5,0)\) .
Find those profits, using \(P(x,y)=8.6x+7.11y\) .
\[P(0,12)=(8.6)(0)+(7.11)(12) = 85.32\]
\[P(3,8)=(8.6)(3)+(7.11)(8) = 82.68\]
\[P(5,0)=(8.6)(5)+(7.11)(0) = 43\]
So, the maximum profit is $85.32.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 12 kilograms and a volume of 10 liters. Each yiv has a mass of 3 kilograms and a volume of 5 liters. The thief can carry a maximum mass of 36 kilograms and a maximum volume of 50 liters. The profit from each xot is $8.48 and the profit from each yiv is $2.95.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
12
3
36
volume (L)
10
5
50
profit ($)
8.48
2.95
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(12x+3y \le 36\)
Volume constraint: \(10x+5y \le 50\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[12x+3y = 36\]
\[10x+5y = 50\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[4x+1y = 12\]
\[2x+1y = 10\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[4x_{\text{int}1}+(1)(0) = 12 \]
\[4x_{\text{int}1} = 12 \]
\[x_{\text{int}1} = \frac{12}{4} = 3 \]
\[2x_{\text{int}2}+(1)(0) = 10 \]
\[2x_{\text{int}2} = 10 \]
\[x_{\text{int}2} = \frac{10}{2} = 5 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(4)(0)+(1)(y_{\text{int1}}) = 12 \]
\[1y_{\text{int}1} = 12 \]
\[y_{\text{int}1} = \frac{12}{1} = 12 \]
\[(2)(0)+(1)y_{\text{int2}} = 10 \]
\[1y_{\text{int}2} = 10 \]
\[y_{\text{int}2} = \frac{10}{1} = 10 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[4x + 1y = 12\]
\[2x + 1y = 10\]
Subtract equations.
\[2x + (0)y = 2\]
Simplify.
\[2x = 2\]
Divide by 2.
\[x = 1\]
Plug \(x = 1\) into the first equation. Solve for \(y\) .
\[(4)(1)+1y = 12\]
\[4+1y = 12\]
\[1y = 8\]
\[y = 8\]
So, the intersection occurs at \((1,\,8)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,10)\) or \((1,8)\) or \((3,0)\) .
Find those profits, using \(P(x,y)=8.48x+2.95y\) .
\[P(0,10)=(8.48)(0)+(2.95)(10) = 29.5\]
\[P(1,8)=(8.48)(1)+(2.95)(8) = 32.08\]
\[P(3,0)=(8.48)(3)+(2.95)(0) = 25.44\]
So, the maximum profit is $32.08.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 15 kilograms and a volume of 14 liters. Each yiv has a mass of 6 kilograms and a volume of 7 liters. The thief can carry a maximum mass of 90 kilograms and a maximum volume of 98 liters. The profit from each xot is $5.47 and the profit from each yiv is $1.21.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
15
6
90
volume (L)
14
7
98
profit ($)
5.47
1.21
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(15x+6y \le 90\)
Volume constraint: \(14x+7y \le 98\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[15x+6y = 90\]
\[14x+7y = 98\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[5x+2y = 30\]
\[2x+1y = 14\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[5x_{\text{int}1}+(2)(0) = 30 \]
\[5x_{\text{int}1} = 30 \]
\[x_{\text{int}1} = \frac{30}{5} = 6 \]
\[2x_{\text{int}2}+(1)(0) = 14 \]
\[2x_{\text{int}2} = 14 \]
\[x_{\text{int}2} = \frac{14}{2} = 7 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(5)(0)+(2)(y_{\text{int1}}) = 30 \]
\[2y_{\text{int}1} = 30 \]
\[y_{\text{int}1} = \frac{30}{2} = 15 \]
\[(2)(0)+(1)y_{\text{int2}} = 14 \]
\[1y_{\text{int}2} = 14 \]
\[y_{\text{int}2} = \frac{14}{1} = 14 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[5x + 2y = 30\]
\[4x + 2y = 28\]
Subtract equations.
\[1x + (0)y = 2\]
Simplify.
\[1x = 2\]
Divide by 1.
\[x = 2\]
Plug \(x = 2\) into the first equation. Solve for \(y\) .
\[(5)(2)+2y = 30\]
\[10+2y = 30\]
\[2y = 20\]
\[y = 10\]
So, the intersection occurs at \((2,\,10)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,14)\) or \((2,10)\) or \((6,0)\) .
Find those profits, using \(P(x,y)=5.47x+1.21y\) .
\[P(0,14)=(5.47)(0)+(1.21)(14) = 16.94\]
\[P(2,10)=(5.47)(2)+(1.21)(10) = 23.04\]
\[P(6,0)=(5.47)(6)+(1.21)(0) = 32.82\]
So, the maximum profit is $32.82.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 5 kilograms and a volume of 9 liters. Each yiv has a mass of 10 kilograms and a volume of 6 liters. The thief can carry a maximum mass of 50 kilograms and a maximum volume of 54 liters. The profit from each xot is $1.86 and the profit from each yiv is $2.47.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
5
10
50
volume (L)
9
6
54
profit ($)
1.86
2.47
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(5x+10y \le 50\)
Volume constraint: \(9x+6y \le 54\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[5x+10y = 50\]
\[9x+6y = 54\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[1x+2y = 10\]
\[3x+2y = 18\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[1x_{\text{int}1}+(2)(0) = 10 \]
\[1x_{\text{int}1} = 10 \]
\[x_{\text{int}1} = \frac{10}{1} = 10 \]
\[3x_{\text{int}2}+(2)(0) = 18 \]
\[3x_{\text{int}2} = 18 \]
\[x_{\text{int}2} = \frac{18}{3} = 6 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(1)(0)+(2)(y_{\text{int1}}) = 10 \]
\[2y_{\text{int}1} = 10 \]
\[y_{\text{int}1} = \frac{10}{2} = 5 \]
\[(3)(0)+(2)y_{\text{int2}} = 18 \]
\[2y_{\text{int}2} = 18 \]
\[y_{\text{int}2} = \frac{18}{2} = 9 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[1x + 2y = 10\]
\[3x + 2y = 18\]
Subtract equations.
\[2x + (0)y = 8\]
Simplify.
\[2x = 8\]
Divide by 2.
\[x = 4\]
Plug \(x = 4\) into the first equation. Solve for \(y\) .
\[(1)(4)+2y = 10\]
\[4+2y = 10\]
\[2y = 6\]
\[y = 3\]
So, the intersection occurs at \((4,\,3)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,5)\) or \((4,3)\) or \((6,0)\) .
Find those profits, using \(P(x,y)=1.86x+2.47y\) .
\[P(0,5)=(1.86)(0)+(2.47)(5) = 12.35\]
\[P(4,3)=(1.86)(4)+(2.47)(3) = 14.85\]
\[P(6,0)=(1.86)(6)+(2.47)(0) = 11.16\]
So, the maximum profit is $14.85.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 15 kilograms and a volume of 16 liters. Each yiv has a mass of 10 kilograms and a volume of 8 liters. The thief can carry a maximum mass of 150 kilograms and a maximum volume of 128 liters. The profit from each xot is $3.98 and the profit from each yiv is $2.21.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
15
10
150
volume (L)
16
8
128
profit ($)
3.98
2.21
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(15x+10y \le 150\)
Volume constraint: \(16x+8y \le 128\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[15x+10y = 150\]
\[16x+8y = 128\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[3x+2y = 30\]
\[2x+1y = 16\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[3x_{\text{int}1}+(2)(0) = 30 \]
\[3x_{\text{int}1} = 30 \]
\[x_{\text{int}1} = \frac{30}{3} = 10 \]
\[2x_{\text{int}2}+(1)(0) = 16 \]
\[2x_{\text{int}2} = 16 \]
\[x_{\text{int}2} = \frac{16}{2} = 8 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(3)(0)+(2)(y_{\text{int1}}) = 30 \]
\[2y_{\text{int}1} = 30 \]
\[y_{\text{int}1} = \frac{30}{2} = 15 \]
\[(2)(0)+(1)y_{\text{int2}} = 16 \]
\[1y_{\text{int}2} = 16 \]
\[y_{\text{int}2} = \frac{16}{1} = 16 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[3x + 2y = 30\]
\[4x + 2y = 32\]
Subtract equations.
\[1x + (0)y = 2\]
Simplify.
\[1x = 2\]
Divide by 1.
\[x = 2\]
Plug \(x = 2\) into the first equation. Solve for \(y\) .
\[(3)(2)+2y = 30\]
\[6+2y = 30\]
\[2y = 24\]
\[y = 12\]
So, the intersection occurs at \((2,\,12)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,15)\) or \((2,12)\) or \((8,0)\) .
Find those profits, using \(P(x,y)=3.98x+2.21y\) .
\[P(0,15)=(3.98)(0)+(2.21)(15) = 33.15\]
\[P(2,12)=(3.98)(2)+(2.21)(12) = 34.48\]
\[P(8,0)=(3.98)(8)+(2.21)(0) = 31.84\]
So, the maximum profit is $34.48.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 14 kilograms and a volume of 10 liters. Each yiv has a mass of 7 kilograms and a volume of 15 liters. The thief can carry a maximum mass of 98 kilograms and a maximum volume of 150 liters. The profit from each xot is $8.3 and the profit from each yiv is $4.36.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
14
7
98
volume (L)
10
15
150
profit ($)
8.30
4.36
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(14x+7y \le 98\)
Volume constraint: \(10x+15y \le 150\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[14x+7y = 98\]
\[10x+15y = 150\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[2x+1y = 14\]
\[2x+3y = 30\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[2x_{\text{int}1}+(1)(0) = 14 \]
\[2x_{\text{int}1} = 14 \]
\[x_{\text{int}1} = \frac{14}{2} = 7 \]
\[2x_{\text{int}2}+(3)(0) = 30 \]
\[2x_{\text{int}2} = 30 \]
\[x_{\text{int}2} = \frac{30}{2} = 15 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(2)(0)+(1)(y_{\text{int1}}) = 14 \]
\[1y_{\text{int}1} = 14 \]
\[y_{\text{int}1} = \frac{14}{1} = 14 \]
\[(2)(0)+(3)y_{\text{int2}} = 30 \]
\[3y_{\text{int}2} = 30 \]
\[y_{\text{int}2} = \frac{30}{3} = 10 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[6x + 3y = 42\]
\[2x + 3y = 30\]
Subtract equations.
\[4x + (0)y = 12\]
Simplify.
\[4x = 12\]
Divide by 4.
\[x = 3\]
Plug \(x = 3\) into the first equation. Solve for \(y\) .
\[(2)(3)+1y = 14\]
\[6+1y = 14\]
\[1y = 8\]
\[y = 8\]
So, the intersection occurs at \((3,\,8)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,10)\) or \((3,8)\) or \((7,0)\) .
Find those profits, using \(P(x,y)=8.3x+4.36y\) .
\[P(0,10)=(8.3)(0)+(4.36)(10) = 43.6\]
\[P(3,8)=(8.3)(3)+(4.36)(8) = 59.78\]
\[P(7,0)=(8.3)(7)+(4.36)(0) = 58.1\]
So, the maximum profit is $59.78.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 20 kilograms and a volume of 12 liters. Each yiv has a mass of 6 kilograms and a volume of 18 liters. The thief can carry a maximum mass of 120 kilograms and a maximum volume of 216 liters. The profit from each xot is $7.53 and the profit from each yiv is $9.91.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
20
6
120
volume (L)
12
18
216
profit ($)
7.53
9.91
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(20x+6y \le 120\)
Volume constraint: \(12x+18y \le 216\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[20x+6y = 120\]
\[12x+18y = 216\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[10x+3y = 60\]
\[2x+3y = 36\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[10x_{\text{int}1}+(3)(0) = 60 \]
\[10x_{\text{int}1} = 60 \]
\[x_{\text{int}1} = \frac{60}{10} = 6 \]
\[2x_{\text{int}2}+(3)(0) = 36 \]
\[2x_{\text{int}2} = 36 \]
\[x_{\text{int}2} = \frac{36}{2} = 18 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(10)(0)+(3)(y_{\text{int1}}) = 60 \]
\[3y_{\text{int}1} = 60 \]
\[y_{\text{int}1} = \frac{60}{3} = 20 \]
\[(2)(0)+(3)y_{\text{int2}} = 36 \]
\[3y_{\text{int}2} = 36 \]
\[y_{\text{int}2} = \frac{36}{3} = 12 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[10x + 3y = 60\]
\[2x + 3y = 36\]
Subtract equations.
\[8x + (0)y = 24\]
Simplify.
\[8x = 24\]
Divide by 8.
\[x = 3\]
Plug \(x = 3\) into the first equation. Solve for \(y\) .
\[(10)(3)+3y = 60\]
\[30+3y = 60\]
\[3y = 30\]
\[y = 10\]
So, the intersection occurs at \((3,\,10)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,12)\) or \((3,10)\) or \((6,0)\) .
Find those profits, using \(P(x,y)=7.53x+9.91y\) .
\[P(0,12)=(7.53)(0)+(9.91)(12) = 118.92\]
\[P(3,10)=(7.53)(3)+(9.91)(10) = 121.69\]
\[P(6,0)=(7.53)(6)+(9.91)(0) = 45.18\]
So, the maximum profit is $121.69.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 4 kilograms and a volume of 18 liters. Each yiv has a mass of 16 kilograms and a volume of 9 liters. The thief can carry a maximum mass of 64 kilograms and a maximum volume of 162 liters. The profit from each xot is $9.09 and the profit from each yiv is $6.88.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
4
16
64
volume (L)
18
9
162
profit ($)
9.09
6.88
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(4x+16y \le 64\)
Volume constraint: \(18x+9y \le 162\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[4x+16y = 64\]
\[18x+9y = 162\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[1x+4y = 16\]
\[2x+1y = 18\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[1x_{\text{int}1}+(4)(0) = 16 \]
\[1x_{\text{int}1} = 16 \]
\[x_{\text{int}1} = \frac{16}{1} = 16 \]
\[2x_{\text{int}2}+(1)(0) = 18 \]
\[2x_{\text{int}2} = 18 \]
\[x_{\text{int}2} = \frac{18}{2} = 9 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(1)(0)+(4)(y_{\text{int1}}) = 16 \]
\[4y_{\text{int}1} = 16 \]
\[y_{\text{int}1} = \frac{16}{4} = 4 \]
\[(2)(0)+(1)y_{\text{int2}} = 18 \]
\[1y_{\text{int}2} = 18 \]
\[y_{\text{int}2} = \frac{18}{1} = 18 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[1x + 4y = 16\]
\[8x + 4y = 72\]
Subtract equations.
\[7x + (0)y = 56\]
Simplify.
\[7x = 56\]
Divide by 7.
\[x = 8\]
Plug \(x = 8\) into the first equation. Solve for \(y\) .
\[(1)(8)+4y = 16\]
\[8+4y = 16\]
\[4y = 8\]
\[y = 2\]
So, the intersection occurs at \((8,\,2)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,4)\) or \((8,2)\) or \((9,0)\) .
Find those profits, using \(P(x,y)=9.09x+6.88y\) .
\[P(0,4)=(9.09)(0)+(6.88)(4) = 27.52\]
\[P(8,2)=(9.09)(8)+(6.88)(2) = 86.48\]
\[P(9,0)=(9.09)(9)+(6.88)(0) = 81.81\]
So, the maximum profit is $86.48.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 12 kilograms and a volume of 15 liters. Each yiv has a mass of 18 kilograms and a volume of 9 liters. The thief can carry a maximum mass of 216 kilograms and a maximum volume of 135 liters. The profit from each xot is $9.1 and the profit from each yiv is $3.07.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
12
18
216
volume (L)
15
9
135
profit ($)
9.10
3.07
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(12x+18y \le 216\)
Volume constraint: \(15x+9y \le 135\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[12x+18y = 216\]
\[15x+9y = 135\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[2x+3y = 36\]
\[5x+3y = 45\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[2x_{\text{int}1}+(3)(0) = 36 \]
\[2x_{\text{int}1} = 36 \]
\[x_{\text{int}1} = \frac{36}{2} = 18 \]
\[5x_{\text{int}2}+(3)(0) = 45 \]
\[5x_{\text{int}2} = 45 \]
\[x_{\text{int}2} = \frac{45}{5} = 9 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(2)(0)+(3)(y_{\text{int1}}) = 36 \]
\[3y_{\text{int}1} = 36 \]
\[y_{\text{int}1} = \frac{36}{3} = 12 \]
\[(5)(0)+(3)y_{\text{int2}} = 45 \]
\[3y_{\text{int}2} = 45 \]
\[y_{\text{int}2} = \frac{45}{3} = 15 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[2x + 3y = 36\]
\[5x + 3y = 45\]
Subtract equations.
\[3x + (0)y = 9\]
Simplify.
\[3x = 9\]
Divide by 3.
\[x = 3\]
Plug \(x = 3\) into the first equation. Solve for \(y\) .
\[(2)(3)+3y = 36\]
\[6+3y = 36\]
\[3y = 30\]
\[y = 10\]
So, the intersection occurs at \((3,\,10)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,12)\) or \((3,10)\) or \((9,0)\) .
Find those profits, using \(P(x,y)=9.1x+3.07y\) .
\[P(0,12)=(9.1)(0)+(3.07)(12) = 36.84\]
\[P(3,10)=(9.1)(3)+(3.07)(10) = 58\]
\[P(9,0)=(9.1)(9)+(3.07)(0) = 81.9\]
So, the maximum profit is $81.90.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 10 kilograms and a volume of 8 liters. Each yiv has a mass of 12 kilograms and a volume of 16 liters. The thief can carry a maximum mass of 120 kilograms and a maximum volume of 128 liters. The profit from each xot is $6.73 and the profit from each yiv is $7.59.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
10
12
120
volume (L)
8
16
128
profit ($)
6.73
7.59
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(10x+12y \le 120\)
Volume constraint: \(8x+16y \le 128\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[10x+12y = 120\]
\[8x+16y = 128\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[5x+6y = 60\]
\[1x+2y = 16\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[5x_{\text{int}1}+(6)(0) = 60 \]
\[5x_{\text{int}1} = 60 \]
\[x_{\text{int}1} = \frac{60}{5} = 12 \]
\[1x_{\text{int}2}+(2)(0) = 16 \]
\[1x_{\text{int}2} = 16 \]
\[x_{\text{int}2} = \frac{16}{1} = 16 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(5)(0)+(6)(y_{\text{int1}}) = 60 \]
\[6y_{\text{int}1} = 60 \]
\[y_{\text{int}1} = \frac{60}{6} = 10 \]
\[(1)(0)+(2)y_{\text{int2}} = 16 \]
\[2y_{\text{int}2} = 16 \]
\[y_{\text{int}2} = \frac{16}{2} = 8 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[5x + 6y = 60\]
\[3x + 6y = 48\]
Subtract equations.
\[2x + (0)y = 12\]
Simplify.
\[2x = 12\]
Divide by 2.
\[x = 6\]
Plug \(x = 6\) into the first equation. Solve for \(y\) .
\[(5)(6)+6y = 60\]
\[30+6y = 60\]
\[6y = 30\]
\[y = 5\]
So, the intersection occurs at \((6,\,5)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,8)\) or \((6,5)\) or \((12,0)\) .
Find those profits, using \(P(x,y)=6.73x+7.59y\) .
\[P(0,8)=(6.73)(0)+(7.59)(8) = 60.72\]
\[P(6,5)=(6.73)(6)+(7.59)(5) = 78.33\]
\[P(12,0)=(6.73)(12)+(7.59)(0) = 80.76\]
So, the maximum profit is $80.76.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
Question prev next
A thief is stealing xots and yivs. Each xot has a mass of 20 kilograms and a volume of 12 liters. Each yiv has a mass of 6 kilograms and a volume of 18 liters. The thief can carry a maximum mass of 120 kilograms and a maximum volume of 216 liters. The profit from each xot is $5.23 and the profit from each yiv is $3.1.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
20
6
120
volume (L)
12
18
216
profit ($)
5.23
3.10
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
Solution Show solution
This is an example of linear programming .
Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.
Mass constraint: \(20x+6y \le 120\)
Volume constraint: \(12x+18y \le 216\)
We can assume that optimal profit will occur at a boundary, so we can just consider the equations.
\[20x+6y = 120\]
\[12x+18y = 216\]
If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:
\[10x+3y = 60\]
\[2x+3y = 36\]
We need to find the \(x\) intercepts and the \(y\) intercepts.
To find the \(x\) -intercepts, set \(y\) to 0.
\[10x_{\text{int}1}+(3)(0) = 60 \]
\[10x_{\text{int}1} = 60 \]
\[x_{\text{int}1} = \frac{60}{10} = 6 \]
\[2x_{\text{int}2}+(3)(0) = 36 \]
\[2x_{\text{int}2} = 36 \]
\[x_{\text{int}2} = \frac{36}{2} = 18 \]
To find the \(y\) -intercepts, set \(x\) to 0.
\[(10)(0)+(3)(y_{\text{int1}}) = 60 \]
\[3y_{\text{int}1} = 60 \]
\[y_{\text{int}1} = \frac{60}{3} = 20 \]
\[(2)(0)+(3)y_{\text{int2}} = 36 \]
\[3y_{\text{int}2} = 36 \]
\[y_{\text{int}2} = \frac{36}{3} = 12 \]
To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.
\[10x + 3y = 60\]
\[2x + 3y = 36\]
Subtract equations.
\[8x + (0)y = 24\]
Simplify.
\[8x = 24\]
Divide by 8.
\[x = 3\]
Plug \(x = 3\) into the first equation. Solve for \(y\) .
\[(10)(3)+3y = 60\]
\[30+3y = 60\]
\[3y = 30\]
\[y = 10\]
So, the intersection occurs at \((3,\,10)\) .
You can plot the boundaries and shade the feasible region .
The maximum profit could occur at \((0,12)\) or \((3,10)\) or \((6,0)\) .
Find those profits, using \(P(x,y)=5.23x+3.1y\) .
\[P(0,12)=(5.23)(0)+(3.1)(12) = 37.2\]
\[P(3,10)=(5.23)(3)+(3.1)(10) = 46.69\]
\[P(6,0)=(5.23)(6)+(3.1)(0) = 31.38\]
So, the maximum profit is $46.69.
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.
